Problem: Recently you have bought a snow walking robot and brought it home. Suppose your home is a cell (0, 0) on an infinite grid.

You also have the sequence of instructions of this robot. It is written as the string s consisting of characters 'L', 'R', 'U' and 'D'. If the robot is in the cell (x, y) right now, he can move to one of the adjacent cells (depending on the current instruction).

  * If the current instruction is 'L', then the robot can move to the left to (x - 1, y); 
  * if the current instruction is 'R', then the robot can move to the right to (x + 1, y); 
  * if the current instruction is 'U', then the robot can move to the top to (x, y + 1); 
  * if the current instruction is 'D', then the robot can move to the bottom to (x, y - 1). 



You've noticed the warning on the last page of the manual: if the robot visits some cell (except (0, 0)) twice then it breaks.

So the sequence of instructions is valid if the robot starts in the cell (0, 0), performs the given instructions, visits no cell other than (0, 0) two or more times and ends the path in the cell (0, 0). Also cell (0, 0) should be visited at most two times: at the beginning and at the end (if the path is empty then it is visited only once). For example, the following sequences of instructions are considered valid: "UD", "RL", "UUURULLDDDDLDDRRUU", and the following are considered invalid: "U" (the endpoint is not (0, 0)) and "UUDD" (the cell (0, 1) is visited twice).

The initial sequence of instructions, however, might be not valid. You don't want your robot to break so you decided to reprogram it in the following way: you will remove some (possibly, all or none) instructions from the initial sequence of instructions, then rearrange the remaining instructions as you wish and turn on your robot to move. 

Your task is to remove as few instructions from the initial sequence as possible and rearrange the remaining ones so that the sequence is valid. Report the valid sequence of the maximum length you can obtain.

Note that you can choose any order of remaining instructions (you don't need to minimize the number of swaps or any other similar metric).

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^4) — the number of test cases.

The next q lines contain test cases. The i-th test case is given as the string s consisting of at least 1 and no more than 10^5 characters 'L', 'R', 'U' and 'D' — the initial sequence of instructions.

It is guaranteed that the sum of |s| (where |s| is the length of s) does not exceed 10^5 over all test cases (∑ |s| ≤ 10^5).

Output

For each test case print the answer on it. In the first line print the maximum number of remaining instructions. In the second line print the valid sequence of remaining instructions t the robot has to perform. The moves are performed from left to right in the order of the printed sequence. If there are several answers, you can print any. If the answer is 0, you are allowed to print an empty line (but you can don't print it).

Example

Input


6
LRU
DURLDRUDRULRDURDDL
LRUDDLRUDRUL
LLLLRRRR
URDUR
LLL


Output


2
LR
14
RUURDDDDLLLUUR
12
ULDDDRRRUULL
2
LR
2
UD
0

Note

There are only two possible answers in the first test case: "LR" and "RL".

The picture corresponding to the second test case:

<image> Note that the direction of traverse does not matter 

Another correct answer to the third test case: "URDDLLLUURDR".

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;



/**
 * Pick questions that you love and solve them with love, try to get AC in one go with cleanest code
 * Just one thing to keep in mind these questions should be out of your reach, which make you step out of comfort zone
 * Try to pick them from different topics
 * CP is a sport, enjoy it. Don't make it pressure cooker or job
 * ****Use pen and paper************* check few examples analyze them, some not given in sample also analyze then code
 * Use pen and paper. Solve on paper then code.
 * If there is some reasoning e.g. sequence/paths, try printing first 100 elements or 100 answers using brute and observe.
 * *********Read question with extreme caution. Mistake is happening here which costs time, WA and easy problem not getting solved.*********
 * Sometimes we make question complex due to misunderstanding.
 * Prefix sum and suffix sum is highly usable concept, look for it.
 * Think of cleanest approach. If too many if else are coming then its indication of WA.
 * Solve 1-2 more questions than you solved during contest.
 */
public class Codeforces {

	public static void main(String args[]) throws IOException {
//		Reader in = new Reader();
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while (t-->0) {
			String s = in.next();
			int l = 0, r = 0, u = 0, d = 0;
			int n = s.length();
			for (int i=0;i<n;i++) {
				char ch = s.charAt(i);
				switch(ch) {
				case 'L':l++;break;
				case 'R':r++;break;
				case 'U':u++;break;
				case 'D':d++;break;
				}
			}
			
			int x = Math.min(l, r);
			int y = Math.min(u, d);
			if (x == 0 && y == 0) {
				System.out.println(0);
				continue;
			}
			if (x == 0) {
				System.out.println(2);
				System.out.println("UD");
				continue;
			}
			if (y == 0) {
				System.out.println(2);
				System.out.println("RL");
				continue;
			}
			StringBuilder sb = new StringBuilder();
			for (int i=0;i<x;i++) sb.append("R");
			for (int i=0;i<y;i++) sb.append("U");
			for (int i=0;i<x;i++) sb.append("L");
			for (int i=0;i<y;i++) sb.append("D");
			String ans = sb.toString();
			System.out.println(ans.length());
			System.out.println(ans);
		}
		in.close();
	}
	
	static class Pair {
		Long x;
		Long y;
		private Pair(long x, long y) {
			this.x = x;
			this.y = y;
		}
		
		@Override
		public boolean equals(Object o) {
			if (o == null || !(o instanceof Pair))
				return false;
			Pair cor = (Pair)o;
			return x.equals(cor.x) && y.equals(cor.y);
		}
		
		@Override
		public int hashCode() {
			int result = 17;
			return result;
		}
		
		static class PairComparatorX implements Comparator<Pair> {

			@Override
			public int compare(Pair o1, Pair o2) {
				return o1.x.compareTo(o2.x);
			}
			
		}

		static class PairComparatorY implements Comparator<Pair> {

			@Override
			public int compare(Pair o1, Pair o2) {
				return o1.y.compareTo(o2.y);
			}
			
		}

	}
	
	private static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;
		StringTokenizer st;
		BufferedReader br;

//		public static void main(String args[]) throws IOException {
//			Reader in = new Reader();
//			int n = in.nextInt();
//			String s[] = new String[n];
//			for (int i=0;i<n;i++) s[i] = in.next();
//			for (int i=0;i<n;i++) System.out.println(s[i]);
//			
//		}
		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}

	}

}